*In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Required number of ways = (8C5 x10C6)
= (8C3 x 10C4)
={(8 x 7 x 6) / ( 3 x 2 x 1)} x {(10 x 9 x 8 x 7) / (4 x 3 x 2 x 1)}
= 11760.
* How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Since each desired number is divisible by 5, so we must have 5 at the unit place.
So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9).
So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits.
So, there are 4 ways of filling it.
So, Required number of numbers = (1 x 5 x 4) = 20.
** In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways = (6C1 x4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3x 4C1) + (6C2)
= (6 x 4) + {(6 x 5)/( 2 x 1) x (4 x 3)/( 2 x 1)} + [{(6 x 5 x 4)/(3 x 2 x 1)}x 4] + (6 x 5)/( 2 x 1)
= (24 + 90 + 80 + 15)
= 209.
* In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?
In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters =7!/ 2!= 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
In 5!/ 3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.
* From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (7C3 x6C2) + (7C4 x 6C1) + (7C5)
= (7 x 6 x 5)/( 3 x 2 x 1) x (6 x 5)/( 2 x 1) + (7C3 x 6C1) + (7C2)
= 525 + [{(7 x 6 x 5)/( 3 x 2 x 1)}x 6] + (7 x 6)/( 2 x 1)
= (525 + 210 + 21)
= 756.
প্রতিটি লেকচারে নতুন নতুন লিখা যুক্ত হচ্ছে, তাই কাঙ্খিত কোন লিখা না পেলে দয়া করে কিছুদিন পর আবার ভিজিট করে দেখবেন।
লিখাতে কিংবা লেকচারে কোন ভুলত্রুটি থাকলে অথবা আপনার কাঙ্খিত লিখা খুঁজে না পেলেইশিখন.কম এর ফ্যানপেইজ অথবা নিচে কমেন্ট কর